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Scenario: you are 1500m away from a building which is 40m high. you have a sniper rifle (in my case, with a muzzle velocity of 854m/s) and you want to shoot an object right at the edge of the 40m high building. with what angle must u adjust your sniper rifle to hit the target?

so this is what i did... a bit diff from the others that have posted.

first i broke my horizontal(x) and vertical(y) components up and got:

using pythagorean theorem:

x = root(729316 -y^2)

y = root(729316 -x^2)

then, since x constant velocity i solved:

x = 1500/t

i subbed x into y eq'n and got:

y = root[729316 - (1500/t)^2]

y = root[(729316t^2 - 2250000)/t^2]

then using my kinematics formula d=v1t + 1/2at^2 i solved for the time it would take the bullet to hit the object using vertical component as my v1.

40 = yt -4.9t^2

i subbed y = root[(729316t^2 - 2250000)/t^2], into the equation:

40 = [(729316t^2 - 2250000)/t^2]t - 4.9t^2

1600 = 729316t^2 - 2250000 - 24.01t^4

24.01t^4 - 729316t^2 + 2251600 = 0

i solved for my realistic roots

t = 1.757 and t = 174.277 (i highly doubt it is the second one though)

now that i know what time equals i solved for the horizontal velocity using

v = d/t

x = 1500/1.757

x = 853.7m/s

then using this horizontal componant and the actual component (854m/s) i solve for the missing angle

853.7 = 854cosr (r = missing angle)

r = 1.52 degrees.

then i found my vertical velocity using y = 854sinr

y = 22.65m/s

then, back to the kinematics eq'n, with all my variables solved for now, i wanted to see if my values would give me my 40m that was required.

d=22.65(1.757) + 0.5(-9.8)(1.757)^2

d=24.67 which is not the desired height.

can someone please tell me where i went wrong here?

i know my explanation is not too great but i will try my best to explain what i am trying to do. any help would be appreciated. thanks

- Tu